Recall
Theorem 32.1. Let be any representation of a finite group (over , or over a field of characteristic ), and suppose is an invariant subspace (i.e., a subrepresentation). Then there exists another invariant subspace such that (as a direct sum of representations).
Corollary 32.2. Any finite-dimensional representation of a finite group decomposes into a direct sum of irreducibles.
Goal: given , find its irreducible representations, and describe how the others decompose into irreducibles.
Lemma 32.3 (Schur’s Lemma).
- If are irreducible representations of , and is any homomorphism of representations, then either or is an isomorphism.
- Over : if is irreducible and is a homomorphism of representations, then is a multiple of the identity.
Proof.
- Given , is an invariant subspace of , i.e. a subrepresentation. Since is irreducible, either ( injective) or (). Similarly, is invariant, hence either zero () or ( surjective). Hence either or is an isomorphism.
- Over , any has an eigenvalue . Then is also equivariant and has nonzero kernel, hence by the above. Thus .
∎
Example 32.4. Let be an irreducible representation of , and , the center of ( commutes with every ). Then the action of satisfies for all , so is equivariant, i.e.
by Schur’s lemma. So acts by a multiple of .
In particular, if is abelian and is irreducible, then every element of acts by a multiple of . This gives another proof that irreducible representations of finite abelian groups are -dimensional.
Representations of
Next we look at the simplest nonabelian group, ( in Fulton–Harris).
- We know the trivial representation (every acts by ).
- There’s another -dimensional representation , coming from the other element of : the alternating representation (also called the sign representation), where acts by .
- We also have the permutation representation with basis , on which acts by permutation matrices: maps .
The permutation representation has an invariant subspace, namely , and we easily find a complementary subrepresentation, namely
This is called the standard representation of ; , and it is irreducible.
What is specific to is that this is the whole story (over ). ( has more irreducible representations; in fact the number of irreducible representations of is , the number of partitions of …)
Proposition 32.5. , and are the only irreducible representations of (over ). Hence any representation of is isomorphic to a direct sum for some (unique) .
Proof. Let be any (finite-dimensional, over ) representation of . Restrict first to the abelian subgroup : let be any -cycle, and any transposition. Then
Restricting the representation to the subgroup generated by (), has a basis of eigenvectors , where with a root of unity. Now let’s see how acts.
If is an eigenvector for , , then
So maps the -eigenspace of to its -eigenspace.
(Remark: for every eigenvector of , is an invariant subspace, since both generators and preserve it. So now we know irreducible representations have .)
Let’s now specialize to the case irreducible, and choose an eigenvector of .
Case : , and by the above, . If is linearly independent of , then satisfies and , so we get an invariant subspace (a trivial subrepresentation) . This contradicts irreducibility. So is a scalar multiple of ; since , . In both cases is invariant, and it is
If is irreducible, this is all of .
Case : then and by the above. Since , these two eigenvectors of are linearly independent; is an invariant subspace, hence by irreducibility it equals . We check that , the standard representation, by mapping to the -eigenvector of in the standard representation (i.e. map to ). ∎
Finding the multiplicities
Given a representation of , , how do we find ?
Answer: Look at the eigenvalues of : the -eigenspace of is , so
whereas the -eigenspaces each have dimension . So the multiplicities of the eigenvalues of give and .
Likewise, acts by on , by on , and by on , so the eigenspaces of have dimension for and for . From this we get , , and .
Example 32.6. Consider the standard representation of , and , also a representation (recall: ). How does decompose into irreducibles?
Show solution
Start with a basis of with
Then has a basis . These are eigenvectors of , with eigenvalues .
Moreover, on the -eigenspace , swaps these two, so is an eigenvector of with eigenvalue . Hence
Similarly, has basis , on which acts by , so
(whereas , perhaps unsurprisingly considering vs. sign).
This generalizes to more complicated groups (we’ll see that eigenvalues go a long way towards classifying representations), but we need some way of organizing the information.
Notes not reached in this lecture, covered at the start of {{LINK:L33}}
Digression: symmetric polynomials
(This is all motivation for the study of characters.)
Observe: an efficient way to store information about (complex) numbers, unordered and possibly with repetitions, is to specify the coefficients of the polynomial of which they are the roots, i.e. . These coefficients are symmetric polynomials in .
acts on the space of polynomials by permuting the variables.
Definition 32.7. A symmetric polynomial is an that is a fixed point of the -action: for all .
Definition 32.8. The elementary symmetric polynomials are
Check: the coefficient of in is, up to the sign , .
Hence: the fundamental theorem of algebra gives a bijection
with inverse
In other terms: the coefficients of the polynomial .
Theorem 32.9. The subring of symmetric polynomials in , i.e. , is isomorphic to the polynomial algebra in variables . I.e., every symmetric polynomial is uniquely a polynomial expression in the elementary symmetric polynomials.
We won’t prove this, but to see why this works, look at the case . The vector space of symmetric polynomials has basis
Observe: any symmetric polynomial in variables can be written as
and we work by induction on degree.